3.4 \(\int (a+b x) \cosh (c+d x) \, dx\)

Optimal. Leaf size=28 \[ \frac{(a+b x) \sinh (c+d x)}{d}-\frac{b \cosh (c+d x)}{d^2} \]

[Out]

-((b*Cosh[c + d*x])/d^2) + ((a + b*x)*Sinh[c + d*x])/d

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Rubi [A]  time = 0.0205861, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3296, 2638} \[ \frac{(a+b x) \sinh (c+d x)}{d}-\frac{b \cosh (c+d x)}{d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*Cosh[c + d*x],x]

[Out]

-((b*Cosh[c + d*x])/d^2) + ((a + b*x)*Sinh[c + d*x])/d

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (a+b x) \cosh (c+d x) \, dx &=\frac{(a+b x) \sinh (c+d x)}{d}-\frac{b \int \sinh (c+d x) \, dx}{d}\\ &=-\frac{b \cosh (c+d x)}{d^2}+\frac{(a+b x) \sinh (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.0513288, size = 27, normalized size = 0.96 \[ \frac{d (a+b x) \sinh (c+d x)-b \cosh (c+d x)}{d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*Cosh[c + d*x],x]

[Out]

(-(b*Cosh[c + d*x]) + d*(a + b*x)*Sinh[c + d*x])/d^2

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Maple [A]  time = 0.009, size = 53, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({\frac{b \left ( \left ( dx+c \right ) \sinh \left ( dx+c \right ) -\cosh \left ( dx+c \right ) \right ) }{d}}-{\frac{cb\sinh \left ( dx+c \right ) }{d}}+a\sinh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*cosh(d*x+c),x)

[Out]

1/d*(b/d*((d*x+c)*sinh(d*x+c)-cosh(d*x+c))-b*c/d*sinh(d*x+c)+a*sinh(d*x+c))

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Maxima [B]  time = 1.13935, size = 92, normalized size = 3.29 \begin{align*} \frac{a e^{\left (d x + c\right )}}{2 \, d} + \frac{{\left (d x e^{c} - e^{c}\right )} b e^{\left (d x\right )}}{2 \, d^{2}} - \frac{{\left (d x + 1\right )} b e^{\left (-d x - c\right )}}{2 \, d^{2}} - \frac{a e^{\left (-d x - c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*cosh(d*x+c),x, algorithm="maxima")

[Out]

1/2*a*e^(d*x + c)/d + 1/2*(d*x*e^c - e^c)*b*e^(d*x)/d^2 - 1/2*(d*x + 1)*b*e^(-d*x - c)/d^2 - 1/2*a*e^(-d*x - c
)/d

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Fricas [A]  time = 1.98283, size = 73, normalized size = 2.61 \begin{align*} -\frac{b \cosh \left (d x + c\right ) -{\left (b d x + a d\right )} \sinh \left (d x + c\right )}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*cosh(d*x+c),x, algorithm="fricas")

[Out]

-(b*cosh(d*x + c) - (b*d*x + a*d)*sinh(d*x + c))/d^2

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Sympy [A]  time = 0.629005, size = 46, normalized size = 1.64 \begin{align*} \begin{cases} \frac{a \sinh{\left (c + d x \right )}}{d} + \frac{b x \sinh{\left (c + d x \right )}}{d} - \frac{b \cosh{\left (c + d x \right )}}{d^{2}} & \text{for}\: d \neq 0 \\\left (a x + \frac{b x^{2}}{2}\right ) \cosh{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*cosh(d*x+c),x)

[Out]

Piecewise((a*sinh(c + d*x)/d + b*x*sinh(c + d*x)/d - b*cosh(c + d*x)/d**2, Ne(d, 0)), ((a*x + b*x**2/2)*cosh(c
), True))

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Giac [A]  time = 1.20289, size = 62, normalized size = 2.21 \begin{align*} \frac{{\left (b d x + a d - b\right )} e^{\left (d x + c\right )}}{2 \, d^{2}} - \frac{{\left (b d x + a d + b\right )} e^{\left (-d x - c\right )}}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*cosh(d*x+c),x, algorithm="giac")

[Out]

1/2*(b*d*x + a*d - b)*e^(d*x + c)/d^2 - 1/2*(b*d*x + a*d + b)*e^(-d*x - c)/d^2